Python, 29 bytes

lambda n:int(bin(n)[:1:-1],2)

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Python, 29 bytes

lambda n:int(bin(n)[:1:-1],2)

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This is an anonymous, unnamed function which returns the result.

First, bin(n) converts the argument to a binary string. We would ordinarily reverse this with the slice notation [::-1]. This reads the string with a step of -1, i.e. backwards. However, binary strings in Python are prefixed with an 0b, and therefore we give the slicing's second argument as 1, telling Python to read backwards terminating at index 1, thus not reading indexes 1 and 0.

Now that we have the backwards binary string, we pass it to int(...) with the second argument as 2. This reads the string as a base 2 integer, which is then implicity returned by the lambda expression.

JavaScript (ES6), 30 bytes

f=(n,q=0)=>n?f(n>>1,q*2+n%2):q

This basically calculates the reverse one bit at a time: We start with q = 0; while n is positive, we multiply q by 2, sever the last bit off of n with n>>1, and add it to q with +n%2. When n reaches 0, the number has been successfully reversed, and we return q.

Thanks to JS's long built-in names, solving this challenge the easy way takes 44 bytes:

n=>+[...n.toString(2),'0b'].reverse().join``

Using recursion and a string, you can get an alternate 32 byte solution:

f=(n,q='0b')=>n?f(n>>1,q+n%2):+q

Java 8, 53 47 46 bytes

-4 bytes thanks to Titus

This is a lambda expression which has the same principle as ETH's answer (although recursion would have been too verbose in Java, so we loop instead):

x->{int t=0;for(;x>0;x/=2)t=t*2+x%2;return t;}

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This can be assigned with IntFunction<Integer> f = ..., and then called with f.apply(num). Expanded, ungolfed and commented, it looks like this:

x -> { 
    int t = 0;           // Initialize result holder   
    while (x > 0) {      // While there are bits left in input:
        t <<= 1;         //   Add a 0 bit at the end of result
        t += x%2;        //   Set it to the last bit of x
        x >>= 1;         //   Hack off the last bit of x
    }              
    return t;            // Return the final result
};

Mathematica, 19 bytes

#~IntegerReverse~2&

J, 6 bytes

|.&.#:

|. reverse

&. under

#: base 2

Jelly, 3 bytes

BUḄ

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B   # convert to binary
 U  # reverse
  Ḅ # convert to decimal

Ruby, 29 28 bytes

->n{("%b"%n).reverse.to_i 2}

"%b" % n formats the input n as a binary string, reverse, then convert back to a number

Usage/Test cases:

m=->n{("%b"%n).reverse.to_i 2}
m[1] #=> 1
m[2] #=> 1
m[3] #=> 3
m[4] #=> 1
m[5] #=> 5
m[6] #=> 3
m[7] #=> 7
m[8] #=> 1
m[9] #=> 9
m[10] #=> 5

C, 48 44 43 42 bytes

-1 byte thanks to gurka and -1 byte thanks to anatolyg:

r;f(n){for(r=n&1;n/=2;r+=r+n%2);return r;}

Previous 44 bytes solution:

r;f(n){r=n&1;while(n/=2)r=2*r+n%2;return r;}

Previous 48 bytes solution:

r;f(n){r=0;while(n)r=2*(r+n%2),n/=2;return r/2;}

Ungolfed and usage:

r;
f(n){
 for(
  r=n&1;
  n/=2;
  r+=r+n%2
 );
 return r;}
}


main() {
#define P(x) printf("%d %d\n",x,f(x))
P(1);
P(2);
P(3);
P(4);
P(5);
P(6);
P(7);
P(8);
P(9);
P(10);
}

05AB1E, 3 bytes

bRC

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Bash/Unix utilities, 24 bytes

dc<<<2i`dc<<<2o$1p|rev`p

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Java 8, 84 79 73 72 bytes

a->a.valueOf(new StringBuffer(a.toString(a,2)).reverse().toString(),2);

Lambda expression. Converts to binary representation, reverses, parses the string in base 2.

-5 bytes thanks to Poke!

-6 more bytes thanks to Poke!

-1, Poke has done more golfing on this than me by now.

Haskell, 36 bytes

0!b=b
a!b=div a 2!(b+b+mod a 2)
(!0)

Same algorithm (and length!) as ETHproductions’ JavaScript answer.

Labyrinth, 89 bytes

2}?_";:_2%}_2/"";{:_2-;;_";!#;@
    "        " "     ;   "  "
    "1_""""""" "1_""""   """"

I'm sure there's a way to reduce loop bytes, but this took long enough to make.

Pyth, 6 bytes

i_.BQ2

Test suite available here.

Explanation

i_.BQ2
    Q     eval(input())
  .B      convert to binary
 _        reverse
i    2    convert from base 2 to base 10

Perl 6, 19 bytes

{:2(.base(2).flip)}

Japt, 5 bytes

¢w n2

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MATL, 4 bytes

BPXB

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Explanation

B     % Input a number implicitly. Convert to binary array
P     % Reverse array
XB    % Convert from binary array to number. Display implicitly

Mathematica, 38 bytes

#+##&~Fold~Reverse[#~IntegerDigits~2]&

Groovy, 46 bytes

{0.parseInt(0.toBinaryString(it).reverse(),2)}

CJam, 8 bytes

ri2bW%2b

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Explanation

ri          e# Read integer
  2b        e# Convert to binary array
    W%      e# Reverse array
      2b    e# Convert from binary array to number. Implicitly display

Batch, 62 bytes

@set/an=%1/2,r=%2+%1%%2
@if %n% gtr 0 %0 %n% %r%*2
@echo %r%

Explanation: On the first pass, %1 contains the input parameter while %2 is empty. We therefore evaluate n as half of %1 and r as +%1 modulo 2 (the % operator has to be doubled to quote it). If n is not zero, we then call ourselves tail recursively passing in n and an expression that gets evaluated on the next pass effectively doubling r each time.

C#, 98 bytes

using System.Linq;using b=System.Convert;a=>b.ToInt64(string.Concat(b.ToString(a,2).Reverse()),2);

PHP, 44 bytes

for(;$n=&$argv[1];$n>>=1)$r+=$r+$n%2;echo$r;

iterative: while input has set bits, pop a bit from input and push it to output. Run with -nr.

reursive, 52 bytes

function r($n,$r=0){return$n?r($n>>1,$r*2+$n%2):$r;}

PHP, 36 bytes

<?=bindec(strrev(decbin($argv[1])));

using builtins: convert to base2, reverse string, convert to decimal
(same as Robert Lerner´s answer, but using a parameter)

R, 55 bytes

sum(2^((length(y<-rev(miscFuncs::bin(scan()))):1)-1)*y)

Reads input from stdin and consequently uses the bin function from the miscFuncs package to convert from decimal to a binary vector.

k, 18 bytes

{2/:|X@&|\X:0b\:x}

Example:

k){2/:|X@&|\X:0b\:x}6
3

Perl 5 (5.12+), 34 bytes

say oct"0b".reverse sprintf"%b",<>

(Input on stdin, output on stdout, requires -E or -M5.012 at no cost).

Straightforward but not all that short. Too bad "reverse" and "sprintf" are such weighty keywords.

oct is an odd legacy name; oct "123" gives the decimal equivalent of octal 123 as you'd expect, but oct "0x123" interprets its input as hex and oct "0b101" interprets it as binary.

Java 7, 94 bytes

Golfed:

int x(int n){return Integer.parseInt(new StringBuffer(Integer.toString(n,2)).reverse()+"",2);}

Ungolfed:

int x(int n)
{
    return Integer.parseInt(new StringBuffer(Integer.toString(n, 2)).reverse() + "", 2);
}

Scala, 40 bytes

i=>BigInt(BigInt(i)toString 2 reverse,2)

Usage:

val f:(Int=>Any)=i=>BigInt(BigInt(i)toString 2 reverse,2)
f(10) //returns 5

Explanation:

i =>          // create an anonymous function with a parameter i
  BigInt(       //return a BigInt contructed from
    BigInt(i)     //i converted to a BigInt
    toString 2    //converted to a binary string
    reverse       //revered
    ,           
    2             //interpreted as a binary string
  )

c/c++ 136 bytes

uint8_t f(uint8_t n){int s=8*sizeof(n)-ceil(log2(n));n=(n&240)>>4|(n&15)<<4;n=(n&204)>>2|(n&51)<<2;n=(n&172)>>1|(n&85)<<1;return(n>>s);}

It's not going to win, but I wanted to take a different approach in c/c++ 120 bytes in the function

#include <math.h>
#include <stdio.h>
#include <stdint.h>

uint8_t f(uint8_t n){
    int s=8*sizeof(n)-ceil(log2(n));
    n=(n&240)>>4|(n&15)<<4;
    n=(n&204)>>2|(n&51)<<2;
    n=(n&172)>>1|(n&85)<<1;
    return (n>>s);
}

int main(){
    printf("%u\n",f(6));
    return 0;
}

To elaborate on what I am doing, I used the log function to determine the number of bits utilized by the input. Than a series of three bit shifts left/right, inside/outside, even/odd which flips the entire integer. Finally a bit shift to shift the number back to the right. Using decimals for bit shifts instead of hex is a pain but it saved a few bytes.